Suppose $v_1,...,v_n \in V$ are nonzero, mutually orthogonal elements
of an inner product space V. Then $v_1,...,v_n$ form an orthogonal
basis for their span W = $span(v_1,...,v_n )\subset V$, which is
therefore a subspace of dimension n = dimW. In particular, if dimV =
n, then $v_1,...,v_n$ form a orthogonal basis for V.

Well they are already orthogonal and they span, so what else do you need to show? If $c_1 v_1 + \cdots + c_n v_n = 0$, consider what happens when you take the inner product of both sides with $v_i$.
–
Michael JoyceOct 23 '12 at 5:59

I have to show that they are linearly independent? So, I must show that $c_i = <u,v_i>$ ?
–
diimensionOct 23 '12 at 6:02

1

If $u = c_1 v_1 + \cdots + c_n v_n$ and if the $v_i$ were orthonormal, then yes, you would show that $c_i = \langle u, v_i \rangle = 0$. But you don't need the $v_i$'s to be orthonormal to prove that they are linearly independent; you just need they are orthogonal to each other.
–
Michael JoyceOct 23 '12 at 6:20

2 Answers
2

Since $v_1, v_2, \ldots, v_n$ is mutually orthogonal, $\{v_1,v_2,\ldots, v_n\}$ is an independent linear system in $V$. Since $dim V=n$ (equal number of elements of $\{v_1,v_2,\ldots, v_n\}$) then $\{v_1,v_2,\ldots, v_n\}$ is a basis of $V$.

To show that $\{v_1,v_2,\ldots, v_n\}$ is independent linear system we consider
$$
a_1v_1+a_2v_2+\ldots+a_nv_n=0,
$$
where $a_i\in \mathbb{R}$. We have
$$
a_1\langle v_1, v_1\rangle + a_2\langle v_1, v_2\rangle+\ldots+a_n\langle v_1, v_n\rangle=0.
$$
Hence $a_1\langle v_1, v_1\rangle=0$ due to the fact that
$$
\langle v_1,v_2\rangle=\langle v_1,v_3\rangle=\ldots=\langle v_1,v_n\rangle=0.
$$. Since $v_1\ne 0$, we have $a_1=0$. Argue similarly we obtain $a_2=a_3=\ldots=a_n=0$.